Solutions 3
Do the following problems from Lang Ex I: 20,21,22,23,25,26,30,39,40,41.
1. Solution to problem 20: We prove that if H is a non-trivial normal subgroup of a p-
group P , then H ∩ Z(P ) is non-trivial. Consider the action of P on H by conjugation.
This action is well-defined as H is normal in P and hence ghg −1 ∈ H for g ∈ P, h ∈ H.
Now, H ∩ Z(P ) is precisely the set of fixed points of this action, and since P is a
p-group we know that |H ∩ Z(P )| must be divisible by p and since, e ∈ H ∩ Z(P ),
|H ∩ Z(P )| ≥ p.
In case that H has order p, it is clear that H ⊂ Z(P ).
2. Solution to problem 21: Clearly PH is a p-group. By Sylow’s theorem, PH is contained
in some p-Sylow subgroup P of G. But, P ∩ H is a subgroup of P , and hence is a
p-group. Thus, P ∩H is a p-subgroup of H containing PH , which is a p-Sylow subgroup
of H. Thus, P ∩ H = PH .
3. Solution to problem 22: Since H is a p-subgroup it is contained in some p-Sylow
subgroup (say P ) of G. But all p-Sylow subgroups are conjugates. Moreover, since H
is normal it is fixed by any conjugation. Thus, H is contained in every other p-Sylow
subgroups as well.
4. Solution to problem 23:
(a) We proved that if H is a p-group and it is contained in the normalizer of a p-Sylow
subgroup P , then H ⊂ P . If H = P 0 is a p-Sylow group, then clearly P 0 = P
because they have the same cardinality.
(b) Since P ⊂ N (P ), and N (P ) = N (P 0 ), then P ⊂ N (P 0 ). Using part (a), this
implies that P = P 0 .
(c) Clearly, N (P ) ⊂ N (N (P )). Now, assume that x ∈ N (N (P )). Hence, xN (P )x−1 =
N (P ). But then, xP x−1 ⊂ xN (P )x−1 = N (P ). But since P is a normal p-Sylow
subgroup of N (P ) it is the unique p-Sylow subgroup of N (P ). But, xP x−1 is a
p-Sylow subgroup of G and hence of N (P ). Hence, xP x−1 = P and x ∈ N (P ).
5. Solution to problem 25:
(a) Using the class equation we proved that any p group has a non-trivial center.
Order of Z cannot be p3 because in that case G is Abelian. Thus, Z has order p
or p2 . If the order of Z is p2 , then G/Z ∼
= Zp and in this case G has to be Abelian.
To see this let ā be a generator for G/Z. Thus, Z, aZ, . . . , ap−1 Z are the cosets of
Z in G and if x, y are arbitrary elements of G, then x = ai z1 , y = aj z2 , for some
o ≤ i, j ≤ p − 1z1 , z2 ∈ Z. But then, xy = yx = ai+j z1 z2 . Hence, order of Z must
be p and Z ∼ = Zp .
Now, G/Z must have order p2 . There are two groups of order p2 , the cyclic group
and the group Zp × Zp (Problem 24). If G/Z is cyclic then the previous argument
will imply again that G is Abelian. Hence, G/Z ∼ = Zp × Zp .
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Solutions 3
(b) Let H be a subgroup of G of order p2 . Let a ∈ Z be a generator of Z. Suppose
a 6∈ H. In this case, H, aH, . . . , ap−1 H are the cosets of H in G. Thus, if x, y ∈ G,
then x = ai h1 , y = aj h2 , for some h1 , h2 ∈ H. But, H being a group of order p2
is Abelian. Hence, xy = ai h1 aj h2 = ai+j h1 h2 = ai+j h2 h1 = aj h2 ai h1 = yx. This
implies that G is Abelian, a contradiction. Hence, a ∈ H and the Z ⊂ H.
Now, G/Z has order p2 and hence is Abelian. Thus, every subgroup of G/Z
is normal. Moreover, there is a correspondence between the subgroups of G/Z
and the subgroups of G containing Z (third isomorphism theorem). Since, H ⊃
Z, it must be normal in G, since under the correspondence normal subgroups
correspond to normal subgroups.
(c) By Sylow’s theorem G contains a subroup of order p2 and by part (b) it contains
Z which is isomorphic to Zp . Now, either H is cyclic or isomorphic to Zp × Zp .
It cannot be the former as every element of G has period ≤ p.
6. Solution to problem 26:
(a) Let G be a group of order pq. The number of p-Sylow subgroups is either 1 or
q. But since q 6= 1 mod p the number of p-Sylow subgroup is 1. Let Hp be the
unique p-Sylow subgroup of G. Now the number of q-Sylow subgroup is either 1
or p. Again since p 6= 1, p < q, the number must be 1 and let Hq be the unique
q-Sylow subgroup. Now, |G \ Hp ∪ Hq | = pq − p − q + 1 > 0. All elements in
G \ Hp ∪ Hq cannot have periods 1, p, q and hence must have period pq and hence
G is cyclic.
(b)
15 = 3 × 5.
Apply, part (a).
7. Solution to problem 30:
(a)
40 = 23 × 5.
The number of 5 Sylow subgroup must be 1 mod 5 and has to divide 40. The
only possible number is 1 and hence it must be normal.
(b)
12 = 22 × 3.
The number of 3-Sylow subgroups is 1 or 4. If it is 1 then it is a normal subgroup.
Else, the number of elements of period 3 is 4 × 2 = 8. Thus, in this case the
number of 2-Sylow subgroups has to be one, making it normal.
8. Solution to problem 39: Let (s1 , . . . , sn−2 ) and (s01 , . . . , s0n−2 ) be two tuples of n − 2
distinct elements of {1, . . . , n}. Let {1, . . . , n} \ {s1 , . . . , sn−2 } = {a, b} and {1, . . . , n} \
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Solutions 3
{s01 , . . . , s0n−2 } = {p, q}. Let σ be the permutation in Sn , defined by
σ(si ) = s0i , 1 ≤ i ≤ n − 2,
σ(a) = p,
σ(b) = q.
If σ is even then we are done. Else, consider the permutation (p, q)σ which also maps
each si to s0i and must be even.
9. Solution to problem 40:
(a) We will assume that n ≥ 5. Let X = {G1 , . . . , Gn } be the set of n cosets of H in
An and assume that H = G1 .
Consider the homomorphism,
φ : An → S x ,
defined by,
φ(σ)(Gi ) = σGi .
Now, since An is simple (for n ≥ 5) ker(φ) is either trivial or An . It cannot be An
because this would imply that H = An . Hence, φ is injective and the image of φ
is a subgroup of SX ∼= Sn isomorphic to An and hence of index 2. We claim that
An is the only subgroup of index 2 in Sn . Suppose there is another subgroup of Sn
(say A) of index 2. Then, A is normal and hence must contain all 3-cycles (refer
to the proof of unsolvability of Sn for n ≥ 5). But since the 3-cycles generate An ,
A has to contain An .
Hence, φ is an isomorphism of An onto AX .
(b) Define ψ : An → An by σGi = Gψ(σ)(i) for 1 ≤ i ≤ n. Clearly, by part (a) ψ is an
automorphism. Moreover, σ ∈ G1 ↔ ψ(σ)(1) = 1 ↔ ψ(σ) ∈ H1 . Thus, ψ is an
automorphism of An mapping H to H1 an ψ −1 maps H1 to H.
Now, suppose that H = Hi for some i. Then, the automorphism, φ : An →
An defined by φ(σ) = (1, i)σ(1, i) maps H1 to Hi . Clearly, if σ fixes 1 then
(1, i)σ(1, i) fixes i. Moreover, φ is induced by an inner automorphism of Sn ,
namely conjugation by (1, i) ∈ Sn .
Now, suppose that H 6= Hi for any i and φ is an automorphism induced by an
inner autmorphism of Sn . Thus, φ(σ) = πσπ −1 for some π ∈ Sn . Let π(1) = i.
Then, φ(σ)(i) = i for all σ ∈ H1 and hence must map H1 to Hi , contradicting the
fact that H 6= Hi for any i.
10. Solution to problem 41:
(a) Consider the 5-sylow subgroups of H. Since, H is simple there must be 6 5-Sylow
subgroups G1 , . . . , G6 and let X = {G1 , . . . , G6 }.
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Solutions 3
Consider the action of H on X by conjugation, which gives an homomorphism
φ : H → SX . Since, H is simple ker(φ) is either trivial or H. It cannot be H
since the action by conjugation is transitive on X. Thus, φ is injective. We now
claim that φ(H) ⊂ AX .
Clearly, since H is simple and φ is injective φ(H) is simple too. Since, AX is a
normal subgroup of SX , φ(H) ∩ AX is a normal subgroup of φ(H) and hence is
trivial or equal to φ(H).
But if φ(H) ∩ AX = {e}, then every non-identity element φ(H) is an odd permu-
tation in SX . Thus, for every σ ∈ φ(H), σ 6= e, σ 2 = e or σ = σ −1 . Moreover,
if σ1 , σ2 are two distinct non-identity elements of φ(H), σ1 σ2 = e implying that
σ1 = σ2−1 = σ2 . Thus in this case, and |φ(H)| ≤ 2, which is not possible. Hence,
φ(H) ⊂ AX .
(b) Follows immediately from the preceding problem.
(c) Since, the action of H on X by conjugation is transitive (since all 5-Sylow sub-
groups are conjugates), it is clear that no element of X is fixed by φ(H). Now,
apply the result of the proved in the preceding problem.
